I got A .57 Hz Need help with these two can\'t seem to get the proper answer Det

Question

I got A

.57  Hz

Need help with these two can't seem to get the proper answer
Determine the amplitude of the resulting motion.
  cm
Determine the phase constant (relative to the original t = 0) of the resulting motion.
  rad

A 1.2 kg block rests on a frictionless surface and is attached to a horizontal spring of constant k = 26 N/m. The block is oscillating with amplitude 15 cm and with phase constant phi = -?/2. A block of mass 0.80 kg is moving from the right at 1.7 m/s. It strikes the first block when the latter is at the rightmost point in its oscillation. The collision is completely inelastic, and the two blocks stick together. Determine the frequency of the resulting motion. Need help with these two can't seem to get the proper answer Determine the amplitude of the resulting motion. Determine the phase constant (relative to the original t = 0) of the resulting motion.

Explanation / Answer

B)

initial momentum of m1=0.8 kg, Pi = m1*v1 = 0.8*1.7 = 1.36 kg.m/s

After collision, let the velocity of the combined mass be v

So, final momentum of the combination , Pf =(m1+m2)*v

By conservation of momentum, Pi = Pf

So, 1.36 = (0.8+1.2)*v

So, v = 0.68 m/s

Now, the whole system(spring and the combined mass), just after the collision has Kinetic energy and stored spring potential energy

So, initial Kinetic energy, KEi = 0.5*(m1+m2)*v^2 = 0.5*(2)*0.68^2 = 0.462 J

Initial Spring potential energy, PEi = 0.5*k*x^2

where x = amplitude of motion before collision = 15 cm = 0.15 m

So, P.Ei = 0.5*26*0.15^2 = 0.293 J

Now, for maximum amplitude(A), the final Kinetic energy , KEf = 0 <---- as it comes to rest momentarily

But all its initial K.E+PE is converted to spring potential energy,

So, KEi + PEi= 0.5*k*A^2

where A = new amplitude of oscillation

So, 0.462 + 0.293 = 0.5*26*A^2

So, A = 0.241m = 24.1 cm <--------answer

C)

The general equation of motion of the combined mass.

x(t) = A*sin(Wt + Q)

where A = amplitude

W = angular frequency = 2*pi*f

Q = phase constant

Initially before collision, Q = -pi/2

initial angular frequency, W = 2*pi*f

where f = sqrt(k/m2)/(2*pi) = sqrt(26/1.2)/(2*pi) = 0.741 Hz

so, at t = 0 , x(0) = x*sin(W*0 - pi/2) = -x

After collision,

for the new t' = 0

at t' = 0, x(0) = A*sin(W*0 - Q) which must be qual to -x

So, -Asin(Q) = -x

So, -24.1*sinQ = -15

So, Q = 0.67 rad <----------answer

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